# Find all pair of consecutive even positive integers, both of the which are larger than 5 such that their sum is less than 23.

$\begin{array}{1 1} (A)\;(7,8),(7,3)\;and\;(2,3) \\(B)\;(6,8),(8,10)\;and\; (10,12) \\(C)\;(5,7),(7,9)\;and \;(2,6) \\(D)\;(2,3),(4,5)\;and\;(3,1) \end{array}$

Toolbox:
• Same Quantity can be added (a subtracted ) to (from ) both sides of the inequality with out changing the sign of the in equality.
• Same positive quantities can be multiplied or divided to both side of the in equality with out changing the sign of the inequality.
• If same negative quantity is multiplied or divided to both sides of the inequality is reversed i.e $'>'$ sign changes to $'<'$ and $'<'$ changes $'>'$ .
Step 1:
Let x and x+2 be two consecutive even positive integers.
Since both the integer are larger then 5. $x > 5$ -----(1)
Also sum of two is less than 23
$x+x+2 < 23$
$=> 2x+2 <23$
$2x < 23-2$
$2x < 21$
Diving by 2 on both sides,
$\large\frac{2x}{2}$$< 21$
$x < \large\frac{21}{2}$
$x < 10.5$-----(2)
Step 2:
Since x is an even positive integer greater than 5 and less then 10.5 x can take value $6,8,10$ .
Thus the required pair of numbers is $( 6,8), (8,10)$ and $(10,12)$
Hence B is the correct answer.